The solution is a little terse, but goes like this (in bold type):

1.Draw line. The rest of the solution depends on this very crucial step, but in my pencil and graph paper model, I can’t figure out how to get a parallel line drawn at that point, I don’t know what to use as a reference length or angle. What I did was build the model in Visio where I’m able to accomplish this and thus proceed with the solution. I use convenient lengths of 8 forDGHparallel toABABand 12 forBC, withAC= 14.7508. In my model, lineDGHhas length 2.6667.

2.Therefore. Amazingly enough, this is true, but I can’t for the life of me figure out or prove why it should be so! The rest of the solution follows, but before I can understand it I need to convince myself of the first two steps. In my model, substituting the lengths in the above equations we obtain 2:6 = 4:12 and 2 = 2 = 2. What I can’t grasp is whyDG:3a=b:3bandDG=a=EADGis equal toa.

3.Therefore. In my model,EF=FGandAF=FD, so thatAF/FD= 1EF=FG= 4.5605 andAF=FD= 5.7963, and of course the ratioAF/FDis 1.

4.AlsoI’ve got 2.6667:8 = 4:12, 2.6667 = 8/3 and 0.6667 = 2.6667 – 2 = 2/3.DH:4a=b:3b,DH=4a/3andGH=DH–DG=a/3.

5.Therefore GC = (1/3)EC and EG = (2/3)EC, and since EF = FG, FC = (2/3)EC.In my model, 4.5605 = (1/3) 13.6815 and 9.1210 = (2/3) 13.6815, and since 4.5605 = 4.5605, 9.1210 = (2/3) 13.6815.

6.Therefore EF/FC = ½.4.5605 / 9.1210 = ½.

7.Therefore (EF/FC) + (AF/FD) = ½ + 1 = 3/2.(4.5605/9.1210) + (5.7963/5.7963) = ½ + 1 = 3/2.